A First Course in Linear Algebra, 2017a

2.2. LU Factorization 103
For a simple illustration of the last claim,
⎡
1 0 0 1 0
⎤
0
⎡
⎣ 0 1 0 0 1 0 ⎦ → ⎣
0 a 1 0 0 1
Now let A be an m × n matrix, say
⎡
A = ⎢
⎣
1 0 0 1 0 0
0 1 0 0 1 0
0 0 1 0 −a 1
⎤
a 11 a 12 ··· a 1n
a 21 a 22 ··· a 2n
⎥
. . . ⎦
a m1 a m2 ··· a mn
and assume A can be row reduced to an upper triangular form using only row operation 3. Thus, in
particular, a 11 ≠ 0. Multiply on the left by E 1 =
⎡
⎤
1 0 ··· 0
− a 21
a 11
1 ··· 0
⎢
⎣
.
. . ..
⎥ . ⎦
− a m1
a 11
0 ··· 1
This is the product of elementary matrices which make modifications in the first column only. It is equivalent
to taking −a 21 /a 11 times the first row and adding to the second. Then taking −a 31 /a 11 times the
first row and adding to the third and so forth. The quotients in the first column of the above matrix are the
multipliers. Thus the result is of the form
⎡
E 1 A = ⎢
⎣
a 11 a 12 ··· a ′ 1n
0 a ′ 22
··· a ′ 2n
. . .
0 a ′ m2
··· a ′ mn
By assumption, a ′ 22 ≠ 0 and so it is possible to use this entry
[
to zero
]
out all the entries below it in the matrix
1 0
on the right by multiplication by a matrix of the form E 2 = where E is an [m − 1]×[m − 1] matrix
0 E
of the form
⎡
⎤
1 0 ··· 0
− a′ 32
a
E =
′ 1 ··· 0
22
⎢ . ⎣ . . .. .
⎥
⎦
0 ··· 1
− a′ m2
a ′ 22
Again, the entries in the first column below the 1 are the multipliers. Continuing this way, zeroing out the
entries below the diagonal entries, finally leads to
E m−1 E n−2 ···E 1 A = U
where U is upper triangular. Each E j has all ones down the main diagonal and is lower triangular. Now
multiply both sides by the inverses of the E j in the reverse order. This yields
A = E −1
1 E−1 2
···E −1
m−1 U
⎤
⎥
⎦
⎤
⎦